01 An introduction to PDEs, part 1

What is a partial differential equation (PDE)?

What are equations? What do we do with them?

• We use equations to describe relationships between quantities.

  • Often, some quantities are known, while others are unknown and need to be determined.

  • We need to “solve for” the unknown quantities in terms of the known ones.

NoteEquations with numbers and algebraic operations

• For example, in the equation \[ 2x + 3 = 9, \] we know the numbers \(2\), \(3\), and \(9\), the algebraic operations connecting them, and we want to solve for the unknown number \(x\).

• The answer is easy: \(x = 3\).

NoteEquations with functions and algebraic operations

• For example, in the equation \[ 2f(x) + 4 = 10x^2, \] we know the numbers \(2\), \(4\), and the function \(10x^2\), the algebraic operations connecting them, and we want to solve for the unknown function \(f(x)\).

• The answer is also easy: \(f(x) = 5x^2 - 2\).

Beyond “algebraic” equations

NoteEquations with functions and differential operations

• For example, in the equation \[ \frac{dy}{dx} = 2y, \] the function \(y(x)\) is unknown. It is embedded in an equation that involves an algebraic operation (multiplication by \(2\)) and a differential operation (taking the derivative with respect to \(x\)).

• The answer is a bit less easy: \(y(x) = Ce^{2x}\), where \(C\) is an arbitrary constant determined by initial (or boundary) conditions.

• Such equations are called differential equations because they involve differential operations.

• This one happens to be an ordinary differential equation (ODE) because the unknown function \(y\) depends on a single variable \(x\).

  • A synonym for “ordinary” (in this context) is “single-variable.”

• ODEs appear everywhere in the real world.

NoteExample: Newton’s second law of motion

This equation says that \[ F = m \frac{d^2x}{dt^2}, \]

where \(x(t)\) is the position of an object as a function of time \(t\), \(m\) is its mass, and \(F\) is the force acting on it.

The central definition

NoteDefinition

A partial differential equation (PDE) is a differential equation that involves an unknown function of multiple variables and its partial derivatives.

• A synonym for “partial” (in this context) is “multi-variable.”

The wave equation in 1 dimension

NoteExample: the \(1\)-dimensional wave equation

This equation says that

\[ \frac{\partial^2 u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}, \]

where \(u(x,t)\) is a two-variable function and \(a\) is a known constant. It models wave-like phenomena, such as vibrations of a string, sound waves, electromagnetic waves, etc.

• \(x\) typically represents a spatial variable (e.g., position along a string).
• \(t\) typically represents time.
• \(u(x,t)\) represents the displacement of the wave at position \(x\) and time \(t\).

import sys

sys.path.append("/Users/johnmyers/dev/johnmyers-phd/teaching/aux-files")
import numpy as np
from plotting import plot_curve_animation, plot_surface

L = 10
blue = "#3399FF"


def u(x, t, n):
    return np.sin(n * np.pi * x / L) * np.cos(n * np.pi * t / L)


def u1(x, t):
    return u(x, t, 1)


x = np.linspace(0, L, 100)
t = np.linspace(0, 2 * L, 100)

plot_curve_animation(
    functions=[u1],
    colors=[blue],
    title="Harmonic n=1",
    x_vals=x,
    y_vals=[-2, 2],
    t_vals=t,
    x_axis_label="x = position",
    y_axis_label="u(x,t) = displacement",
)

X, T = np.meshgrid(x, t)

plot_surface(
    function=u1,
    X=X,
    Y=T,
    x_axis_label="x = position",
    y_axis_label="t = time",
    z_axis_label="u(x,t) = displacement",
    title="Harmonic n=1 over space and time",
)

def u2(x, t):
    return u(x, t, 2)


plot_curve_animation(
    [u2],
    labels=["Harmonic n=2"],
    colors=[blue],
    title="Harmonic n=2",
    x_vals=x,
    y_vals=[-2, 2],
    t_vals=t,
    x_axis_label="x = position",
    y_axis_label="u(x,t) = displacement",
)

plot_surface(
    function=u2,
    X=X,
    Y=T,
    x_axis_label="x = position",
    y_axis_label="t = time",
    z_axis_label="u(x,t) = displacement",
    title="Harmonic n=2 over space and time",
)

def u3(x, t):
    return u(x, t, 3)


plot_curve_animation(
    [u3],
    labels=["Harmonic n=3"],
    colors=[blue],
    title="Harmonic n=3",
    x_vals=x,
    y_vals=[-2, 2],
    t_vals=t,
    x_axis_label="x = position",
    y_axis_label="u(x,t) = displacement",
)

plot_surface(
    function=u3,
    X=X,
    Y=T,
    x_axis_label="x = position",
    y_axis_label="t = time",
    z_axis_label="u(x,t) = displacement",
    title="Harmonic n=3 over space and time",
)

def superposition(x, t):
    return u2(x, t) + u3(x, t)


plot_curve_animation(
    [u2, u3, superposition],
    labels=["n=2", "n=3", "superposition"],
    colors=[blue, "#FF3399", "#33FF99"],
    title="Superposition of harmonics n=2 and n=3",
    x_vals=x,
    y_vals=[-2, 2],
    t_vals=t,
    x_axis_label="x = position",
    y_axis_label="u(x,t) = displacement"
)

plot_surface(
    function=superposition,
    X=X,
    Y=T,
    x_axis_label="x = position",
    y_axis_label="t = time",
    z_axis_label="u(x,t) = displacement",
    title="Superposition of harmonics n=2 and n=3 over space and time",
)

The wave equation in 2 dimensions

NoteExample: the \(2\)-dimensional wave equation

This equation says that

\[ \frac{\partial^2 u}{\partial t^2} = a^2 \left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right), \]

where \(u(x,y,t)\) is a three-variable function and \(a\) is a known constant. It models wave-like phenomena in two spatial dimensions, such as waves on the surface of a pond.

• \(x\) and \(y\) typically represent spatial variables (e.g., position on a surface).
• \(t\) typically represents time.
• \(u(x,y,t)\) represents the displacement of the wave at position \((x,y)\) and time \(t\).

import numpy as np
from plotting import plot_surface_animation


def u(x, y, t):
    return (
        np.sin(2 * np.pi * x) * np.sin(3 * np.pi * y) * np.sin(np.sqrt(13) * np.pi * t)
    )


def u0(x, y):
    return u(x, y, 0)


x = np.linspace(0, 1, 50)
y = np.linspace(0, 1, 50)
t_vals = np.linspace(0, 2, 200)
X, Y = np.meshgrid(x, y)

plot_surface_animation(
    function=u,
    init_function=u0,
    X=X,
    Y=Y,
    z_range=[-1, 1],
    t_vals=t_vals,
    x_axis_label="x = position",
    y_axis_label="y = position",
    z_axis_label="u(x,y,t) = displacement",
    title="Solution of 2D wave equation",
)

def superposition(x, y, t):
    return np.sin(2 * np.pi * x) * np.sin(3 * np.pi * y) * np.sin(
        np.sqrt(13) * np.pi * t
    ) + np.sin(3 * np.pi * x) * np.sin(4 * np.pi * y) * np.sin(np.sqrt(25) * np.pi * t)


def sup0(x, y):
    return superposition(x, y, 0)


plot_surface_animation(
    function=superposition,
    init_function=sup0,
    X=X,
    Y=Y,
    z_range=[-2, 2],
    t_vals=t_vals,
    x_axis_label="x = position",
    y_axis_label="y = position",
    z_axis_label="u(x,y,t) = displacement",
    title="Superposition of 2D wave equation solutions",
)

The heat equation in 1 dimension

NoteExample: the \(1\)-dimensional heat equation

This equation says that

\[ \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}, \]

where \(u(x,t)\) is a two-variable function and \(k\) is a known constant. It models “stationary” diffusion processes (i.e., no “drift”), such as the diffusion of heat in a rod over time.

• \(x\) typically represents a spatial variable (e.g., position along a rod).
• \(t\) typically represents time.
• \(u(x,t)\) represents the temperature at position \(x\) and time \(t\).

blue = "#3399FF"
pink = "#FF3399"
green = "#33FF99"


def u3(x, t):
    return np.exp(-(9 * np.pi**2 * t) / 4) * np.cos((3 * np.pi * x) / 2)


def u4(x, t):
    return np.exp(-(4**2 * np.pi**2 * t) / 4) * np.sin((4 * np.pi * x) / 2)


def superposition(x, t):
    return u3(x, t) + u4(x, t)


x_vals = np.linspace(0, 2, 100)
y_vals = [-2, 2]
t_vals = np.linspace(0, 1, 100)
x_axis_label = "x = position"
y_axis_label = "u(x, t) = temperature"

plot_curve_animation(
    functions=[u3, u4, superposition],
    x_vals=x_vals,
    y_vals=y_vals,
    t_vals=t_vals,
    x_axis_label=x_axis_label,
    y_axis_label=y_axis_label,
    title="Solutions of 1D heat equation",
    colors=[blue, pink, green],
    labels=["u₃", "u₄", "u₃ + u₄"],
)

t_vals = np.linspace(0, 0.2, 100)
X, T = np.meshgrid(x_vals, t_vals)
plot_surface(
    function=superposition,
    X=X,
    Y=T,
    x_axis_label="x = position",
    y_axis_label="t = time",
    z_axis_label="u(x,t) = temperature",
    title="A solution of the 1D heat equation over space and time",
    aspectmode="manual",
    aspectratio=dict(x=1, y=0.5, z=0.75),
)

The Fokker-Planck (diffusion) equation

NoteExample: the Fokker-Planck (diffusion) equation

This equation says that

\[ \frac{\partial u}{\partial t} = -\mu \frac{\partial u}{\partial x} + \frac{\sigma^2}{2} \frac{\partial^2u}{\partial x^2}, \]

where \(u(x,t)\) is a two-variable function and \(\mu\) and \(\sigma\) are known constants. It models diffusion processes with “drift” or “trend,” such as the evolution of probability densities in random processes. A good example to have in mind is the random price of a stock or some financial asset over time:

• \(x\) then represents the stock price.
• \(t\) represents time.
• \(\mu\) represents the trend of the stock, with \(\mu>0\) indicating an average upward trend and \(\mu<0\) indicating an average downward trend.
• \(\sigma\) represents the volatility (i.e., randomness) of the stock price.
• \(u(x,t)\) represents the probability density of the stock price being \(x\) at time \(t\).

blue = "#3399FF"
pink = "#FF3399"
green = "#33FF99"

sigma = 1
mu = 1
x0 = 1


def p(x, t):
    return (1 / (sigma**2 * np.sqrt(2 * np.pi * t))) * np.exp(
        -((x - x0 - mu * t) ** 2) / (2 * sigma**2 * t)
    )


x_vals = np.linspace(-1, 12, 200)
y_vals = [-0.1, 1.4]
t_vals = np.linspace(0.1, 10, 100)
x_axis_label = "x = stock price"
y_axis_label = "u(x, t) = probability density"

plot_curve_animation(
    functions=[p],
    x_vals=x_vals,
    y_vals=y_vals,
    t_vals=t_vals,
    x_axis_label=x_axis_label,
    y_axis_label=y_axis_label,
    title="A solution to the diffusion equation",
    colors=[blue],
)

X, T = np.meshgrid(x_vals, t_vals)
plot_surface(
    function=p,
    X=X,
    Y=T,
    x_axis_label=x_axis_label,
    y_axis_label="t = time",
    z_axis_label=y_axis_label,
    title="A solution of the diffusion equation over space and time",
    cmin=0,
    cmax=0.2,
)

From beads to waves: deriving the wave equation

Deriving the wave equation (part 1)

• We imagine that we have a vibrating string of length \(L\) fixed at both ends.

  • For definiteness, suppose that the string lies along the \(x\)-axis, and is clamped at \(x=0\) and \(x=L\).

NoteThe wave problem as a boundary-value/initial-value problem

• Now, suppose you know:

  • The initial displacement of the string, i.e., you know the function \(u(x,0)\) for \(0 \leq x \leq L\).

  • The initial velocity of each point on the string, i.e., you know the derivative \(\displaystyle\frac{\partial u}{\partial t}(x,0)\) for \(0 \leq x \leq L\).

  • The string is fixed at both ends, so \(u(0,t) = 0\) and \(u(L,t) = 0\) for all \(t \geq 0\).

• From these boundary and initial conditions, can we determine how the string moves over time? In other words, can we find \(u(x,t)\) for all \(x\) and \(t\)?

• The answer is yes, provided that we are willing to make some simplifying assumptions about the string and its motion.

• Essentially, the goal is to derive an equation that \(u(x,t)\) must satisfy. This will be the wave equation.

• We follow an argument begun by John Bernoulli in 1727, and later advanced by d’Alembert in 1747.

Deriving the wave equation (part 2)

NoteDeriving the wave equation via beads

• Here’s the plan:

  1. Suppose that the string is chopped up into \(n\) equal length segments, with a small bead between each segment. The beads can move up and down, but not left and right. The beads all have equal mass. There are \(n+1\) beads in total, including the two fixed beads at the ends.

  2. The \(n\) string segments are weightless, and perfectly flexible and elastic. In particular, the segments are always straight and do not curve. (You could think of the string segments as tiny springs.)

  3. We analyze the motion of each of the \(n-1\) interior beads using Newton’s laws of motion. We obtain equations of motion for each bead.

  4. Then, to treat the “real, continuous” string, we let the number \(n\) of beads tend to \(\infty\), while the mass of each bead also tends to \(0\).

  5. As \(n\to \infty\), the equations derived in step 2 will combine to reveal the wave equation.

Deriving the wave equation (part 3)

• The \(x\)-positions of the beads are given by \[ x_k = k\frac{L}{n}, \quad k = 1, 2, \ldots, n, \] including the one at the right end \(x=L\) (which doesn’t move).

• The vertical displacement of the \(k\)-th bead at time \(t\) is given by \(u_k\).

• Each of the interior beads feels a pull (i.e., tension force) from the bead to its left and a pull from the bead to its right.

  • We assume that all these tension forces have the same magnitude \(T\) (a known constant).

  • We assume that the mass of each bead is \(m\) (also a known constant).

• By Newton’s second law, the vertical acceleration of the \(k\)-th bead is given by \[ m \frac{d^2 u_k}{dt^2} = F_{k}, \] where \(F_{k}\) is the net vertical force acting on the \(k\)-th bead. Since the beads do not move left or right, the horizontal components of the tension forces cancel out.

• If \(\theta_{k+1}\) is the angle that the string to the right of the \(k\)-th bead makes with the positive \(x\)-axis, measured counterclockwise, and if \(\theta_{k-1}\) is the angle that the string to the left of the \(k\)-th bead makes with the negative \(x\)-axis, measured clockwise, then a little trigonometry shows that \[ F_{k} = T \sin\theta_{k+1} + T \sin\theta_{k-1}. \]

• But as long as these angles are small, we have \[ \sin\theta_{k+1} \approx \tan \theta_{k+1} = \frac{u_{k+1} - u_k}{L/n} \quad \text{and} \quad \sin\theta_{k-1} \approx \tan \theta_{k-1} = \frac{u_{k-1} - u_k}{L/n}. \]

Deriving the wave equation (part 4)

• Thus, for small angles, we have \[ F_{k} = \frac{Tn}{L} (u_{k+1} - 2u_k + u_{k-1}) \] where we’ve written “\(=\)” for simplicity instead of “\(\approx\)”.

• If we write \(M = nm\) (which is almost the total mass of the string—why?), then Newton’s second law becomes \[ \frac{d^2 u_k}{dt^2} = \frac{Tn^2}{ML} (u_{k+1} - 2u_k + u_{k-1}). \]

• Set \(a^2 = \frac{TL}{M}\) and \(\Delta x = \frac{L}{n}\), and then rearrange to get \[ \frac{d^2 u_k}{dt^2} = a^2 \left(\frac{u_{k+1} - 2u_k + u_{k-1}}{(\Delta x)^2}\right). \]

• Now, switch to function notation by writing \(u(x,t)\) instead of \(u_k\), and \(x\) instead of \(x_k\). Then the equation becomes \[ \frac{\partial^2 u(x, t)}{\partial t^2} = a^2 \left(\frac{u(x + \Delta x, t) - 2u(x, t) + u(x - \Delta x, t)}{(\Delta x)^2}\right). \]

Deriving the wave equation (part 5)

• From calculus, we know that \[ u(x+\Delta x, t) = u(x,t) + \frac{\partial u}{\partial x}(x,t)\Delta x + \frac{\partial^2 u}{\partial x^2}(x,t)\frac{(\Delta x)^2}{2} + (\text{higher order terms in } \Delta x), \] and \[ u(x-\Delta x, t) = u(x,t) - \frac{\partial u}{\partial x}(x,t)\Delta x + \frac{\partial^2 u}{\partial x^2}(x,t)\frac{(\Delta x)^2}{2} + (\text{higher order terms in } \Delta x). \]

• Thus \[ u(x + \Delta x, t) - 2u(x, t) + u(x - \Delta x, t) = \frac{\partial^2 u}{\partial x^2}(x,t)(\Delta x)^2 + (\text{higher order terms in } \Delta x). \]

• And so \[ \frac{\partial^2 u(x, t)}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}(x,t) + (\text{terms that vanish as } \Delta x \to 0). \]

• If we remember that \(M=mn\) and \(\Delta x = L/n\), then we want \(n\to \infty\) and \(m\to 0\) in such a way that \(M\) remains constant. Then \(\Delta x \to 0\) as \(n\to \infty\), while \(a^2 = \frac{TL}{M}\) remains constant.

• This gives the wave equation: \[ \frac{\partial^2 u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}. \] Here, \(a^2 = T/ \rho\), where \(\rho\) is the mass per unit length of the string and \(T\) is the tension force.